How to see that this definition of Cantor Lebesgue function is equivalent I refer to the following alternative definition of the Cantor function from Wikipedia:
See figure. To formally define the Cantor function c : [0,1] → [0,1], let x be in [0,1] and obtain c(x) by the following steps:
1) Express x in base 3.
2) If x contains a 1, replace every digit after the first 1 by 0.
3) Replace all 2s with 1s.
4) Interpret the result as a binary number. The result is c(x).
I find this definition nice and explicit. Sure, it works when I plug in some numbers, but I have no intuition on why it works? Thanks for any help.
Basically, I am trying to see why this definition is equivalent to the other one in my textbook:

 A: I don't think this is an easy problem, because the first definition is an explicit function $f$, while the second is a limit of a sequence of functions. See here  and observe that the limit function $F$ is not even found explicitly either; rather, it arises from a Cauchy sequence, using completeness of $[0,1]$. 
But for example, if we look at $F_{n}(x)=\frac{3^{n}}{2^{n}}x;\ x\in [0,\frac{1}{3^{n}}]$, then 
$F_n(\frac{1}{3^{n}})=\frac{1}{2^{n}}=\underbrace{.00\cdots 0  }_{\text{n-1}}100\cdots $ in its binary expansion. 
On the other hand, if we write $\frac{1}{3^{n}}=\underbrace{.00\cdots 0  }_{\text{n}}2222\cdots $, in its ternary expansion then using the explicit formula, we also get $\underbrace{.00\cdots 0  }_{\text{n}}1111\cdots = \underbrace{.00\cdots 0  }_{\text{n-1}}1000\cdots$ in the binary expansion.
It's easy to show that if $m>n$ then $F_m$ and $F_n$ agree at $\frac{1}{3^{n}}$,so the limit $F$ is equal to $f$ at these points.  
Obviously, this isn't a proof, but it indicates how you might proceed. In fact, if you look at the definition of the $F_n$, I think you can show, without loss of generality, that it suffices to consider the intervals $[0,1/3^n]$ when doing the analysis because, at the $n^{th}$ stage, there are $2^{n-1}$ intervals of equal length, all obtained by trisecting the intervals from the $(n-1)^{th}$ stage. 
